As shown in the figure (Figure 1) , a layer of water covers a slab of material X

Question

As shown in the figure (Figure 1) , a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the path indicated. Assume that?1 is a1 = 46? and that ?2 is a2 = 66.78?.

Find the index of refraction of material X . Use 1.34 for the index of refraction of water.

Find the angle the light makes with the normal in the air.

What is the critical angle for substance X if the light was leaving it and going out to air?

What is the speed of light in substance X?

Explanation / Answer

1)

Using snell's law

nx * sin(theta1) = nw * sin(theta2)

nx * sin(66.78) = 1.34 *sin(46)

nx = 1.05

the refractive index of X is 1.05

2)

Now, using snell's law

n1* sin(theta(air)) = nw * sin(46)

1 * sin(theta(air)) = 1.34 * sin*(46)

theta(air) = 76.6 degree

3)

for critical angle

nx * sin(theta(critical)) = na * sin(90)

1.05 * sin(theta(critical)) = 1

theta(critical) = 72.2 degree

4)

speed of light in X = 3 *10^8/1.05

speed of light in X = 2.857 *10^8 m/s

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